Math Trailhead

\(42\) and \(49\) have a common factor of \(7\text{.}\)

\(x^{2}\) and \(x^{1}\) have a common factor of \(x^{1}\)

\(28\) and \(7\) have a common factor of \(7\text{.}\)

\(x^{5}\) and \(x^{4}\) have a common factor of \(x^{4}\)

So, \({28x^{5}-7x^{4}} \rightarrow {7x^{4}\mathopen{}\left(4x-1\right)}\text{.}\)

This problem has a greatest common factor (GCF) of \(y\text{.}\) If we factor the GCF out, we have

\(y(20y^2 - 9y - 20)\)

Now we can use the AC-method to factor the remaining trinomial. We are looking for two numbers with a product of \(20\cdot-20=-400\) and a sum of \(-9\text{.}\) Those numbers are \(-25\) and \(16\text{.}\)

We can use those two numbers to rewrite the middle term as two terms:

\(y(20y^2 - 25y + 16y - 20)\)

Now we can factor by grouping

\(y(5y( 4y - 5 ) + 4( 4y - 5 ))\)

\({y\mathopen{}\left(4y-5\right)\mathopen{}\left(5y+4\right)}\)

This problem has a greatest common factor (GCF) of \(10z\text{.}\) If we factor the GCF out, we have

\(10z(10z^2+13z + 4)\)

Now we can use the AC-method to factor the remaining trinomial. We are looking for two numbers with a product of \(10\cdot4=40\) and a sum of \(13\text{.}\) Those numbers are \(5\) and \(8\text{.}\)

We can use those two numbers to rewrite the middle term as two terms:

\(10z(10z^2 + 5z + 8z + 4)\)

Now we can factor by grouping

\(10z(5z( 2z + 1 ) + 4( 2z + 1 ))\)

\({10z\mathopen{}\left(2z+1\right)\mathopen{}\left(5z+4\right)}\)

There is a greatest common factor of \(5\text{.}\) Start by factoring that out.

\(5({x^{3}-6x^{2}+2x-12})\)

Then we can factor the remaining expression by grouping.

\(5(x^2(x - 6) + 2(x - 6))\)

\({5\mathopen{}\left(x^{2}+2\right)\mathopen{}\left(x-6\right)}\)

\(3({y+1})\) and \(4({y+1})\) have a common factor of \({y+1}\text{.}\)

Think about it. \({y+1}\) will be the same value in each term, no matter what value \(y\) represents.

Each term will be a multiple of \({y+1}\text{,}\) so it is a common factor.

So, \({3x\mathopen{}\left(y+1\right)-4\mathopen{}\left(y+1\right)} \rightarrow {\left(y+1\right)\mathopen{}\left(3x-4\right)}\text{.}\)

Begin by grouping the first two terms and last two terms:

We are allowed to do this by the associative property of addition.

\((12 A B + 20 A) + (21 B + 35)\)

Factor each group separately:

\((12 A B + 20 A) \rightarrow 4 A ( 3 B + 5 )\)

\((21 B + 35) \rightarrow 7 ( 3 B + 5 )\)

Each term is a multiple of \(3 B + 5\text{,}\) so it is a common factor.

So, \({12AB+20A+21B+35} \rightarrow 4 A ( 3 B + 5 ) + 7 ( 3 B + 5 )\)

and \(4 A ( 3 B + 5 ) + 7 ( 3 B + 5 ) \rightarrow {\left(3B+5\right)\mathopen{}\left(4A+7\right)}\)

Begin by grouping the first two terms and last two terms:

You must be careful, subtraction is not associative like addition is.

\(-30 A B + 35 A - 18 B + 21\)

\((-30 A B + 35 A) + (-18 B + 21)\)

Factor each group separately:

\((-30 A B + 35 A) \rightarrow 5 A ( -6 B + 7 )\)

\((-18 B + 21) \rightarrow 3 ( -6 B + 7 )\)

Each term is a multiple of \(-6 B + 7\text{,}\) so it is a common factor.

So, \({-30AB+35A-18B+21} \rightarrow 5 A ( -6 B + 7 ) + 3 ( -6 B + 7 )\)

and \(5 A ( -6 B + 7 ) + 3 ( -6 B + 7 ) \rightarrow {\left(-6B+7\right)\mathopen{}\left(5A+3\right)}\)

There are a few ways to approach this type of problem- we will demonstrate the factor by grouping method, but it can be done in other ways

\(\begin{aligned} {10x^{2}+80xy+3xy+24y^{2}}\amp =10x(x+8y)+3y(x+8y) \\ \amp ={\left(10x+3y\right)\mathopen{}\left(x+8y\right)} \end{aligned}\)

Note that this answer can be checked by using the FOIL (First Outside Inside Last) technique (exercise).

Let’s follow the hint and begin by factoring each pair of terms first- this should help to highlight the pattern.

\(\begin{aligned} {35x^{5}+105x^{4}+14x^{4}+42x^{3}+28x^{3}+84x^{2}}\amp =({35x^{5}+105x^{4}})+({14x^{4}+42x^{3}})+({28x^{3}+84x^{2}})\\ \amp ={35x^{4}\mathopen{}\left(x+3\right)+14x^{3}\mathopen{}\left(x+3\right)+28x^{2}\mathopen{}\left(x+3\right)}\\ \amp ={7x^{2}\mathopen{}\left(x+3\right)\mathopen{}\left(5x^{2}+2x+4\right)} \end{aligned}\)